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Revision as of 18:52, 7 May 2005 editOleg Alexandrov (talk \| contribs)Administrators47,244 edits revert vandalism← Previous edit		Revision as of 20:48, 7 May 2005 edit undoOleg Alexandrov (talk \| contribs)Administrators47,244 edits →Explanation: bit of rewordingNext edit →
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	== Explanation ==		== Explanation ==

	The key step to understand here is that the infinite geometric series is convergent.		The key step to understand here is that the following infinite geometric series is convergent:

	:<math>\sum_{k=0}^\infty \left( \frac{1}{10} \right)^k = \frac{1}{1 - \frac{1}{10}}.</math>		:<math>\sum_{k=0}^\infty \left( \frac{1}{10} \right)^k = \frac{1}{1 - \frac{1}{10}}.</math>

Revision as of 20:48, 7 May 2005

In mathematics, one could easily fall in the trap of thinking that while 0.999... is certainly close to 1, nevertheless the two are not equal. Here's a proof that they actually are.

Proof

$0.999\ldots$	$={\frac {9}{10}}+{\frac {9}{100}}+{\frac {9}{1000}}+\cdots$
	$=-9+{\frac {9}{1}}+{\frac {9}{10}}+{\frac {9}{100}}+{\frac {9}{1000}}+\cdots$
	$=-9+9\times \sum _{k=0}^{\infty }\left({\frac {1}{10}}\right)^{k}$
	$=-9+9\times {\frac {1}{1-{\frac {1}{10}}}}$
	$=1.\,$

Explanation

The key step to understand here is that the following infinite geometric series is convergent:

\sum _{k=0}^{\infty }\left({\frac {1}{10}}\right)^{k}={\frac {1}{1-{\frac {1}{10}}}}.

See also

External proofs

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Proofs